# How do you integrate (e^x)sin4x dx?

Apr 29, 2015

This integral is a cyclic one and has to be done two times with the theorem of the integration by parts, that says:

$\int f \left(x\right) g ' \left(x\right) \mathrm{dx} = f \left(x\right) g \left(x\right) - \int g \left(x\right) f ' \left(x\right) \mathrm{dx}$

We can assume that $f \left(x\right) = \sin 4 x$ and $g ' \left(x\right) \mathrm{dx} = {e}^{x} \mathrm{dx}$,

$f ' \left(x\right) \mathrm{dx} = \cos 4 x \cdot 4 \mathrm{dx}$

$g \left(x\right) = {e}^{x}$,

so:

$I = {e}^{x} \sin 4 x - \int {e}^{x} \cos 4 x \cdot 4 \mathrm{dx} =$

$= {e}^{x} \sin 4 x - 4 \int {e}^{x} \cos 4 x \mathrm{dx} = \left(1\right)$.

And now...again:

$f \left(x\right) = \cos 4 x$ and $g ' \left(x\right) \mathrm{dx} = {e}^{x} \mathrm{dx}$,

$f ' \left(x\right) \mathrm{dx} = - \sin 4 x \cdot 4 \mathrm{dx}$

$g \left(x\right) = {e}^{x}$.

So:

$\left(1\right) = {e}^{x} \sin 4 x - 4 \left[{e}^{x} \cos 4 x - \int {e}^{x} \left(- \sin 4 x \cdot 4\right) \mathrm{dx}\right] =$

$= {e}^{x} \sin 4 x - 4 {e}^{x} \cos 4 x - 16 \int {e}^{x} \sin 4 x \mathrm{dx}$.

So, finally, we can write:

$I = {e}^{x} \sin 4 x - 4 {e}^{x} \cos 4 x - 16 I \Rightarrow$

$17 I = {e}^{x} \sin 4 x - 4 {e}^{x} \cos 4 x \Rightarrow$

$I = {e}^{x} / 17 \left(\sin 4 x - 4 \cos 4 x\right) + c$.